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Current I is coming from the Z axis and then it flows out uniformly towards the circumference of this disk. The disk has thickness t and Radius R. There is also a constant B field in the Z direction(going up). I am asked to find the torque exerted on the disk. Here is my attempt: I assume an angle $d\phi$ as shown in the picture and imagine that the current flows through the arc corresponding to that angle. The area of the arc with the thickness t is $A=Rd\phi t $

Now I will find $dI$ using the fact that we have uniform distribution

$J=I/A=>dI=JdA=JRd \phi t= \frac{IRd\phi t}{R2 \pi t}=>$ $$dI=\frac{Id\phi}{2\pi}$$ Now I can write $dF$ in terms of $\phi$:

$$dF=dI(R \times B)=\frac{Id\phi}{2\pi}RB$$ $$F=\int_0^πdF=BIR/2$$ I integrated from 0 to π because of the symmetry and since the forces are couples $$T=(F)(2R)=-BIR^2\hat z$$ My notes say that the result should be $\frac{-BIR^2}{2}\hat z$. And it seems I oversimplified the problem. The solution from the notes also assumes a small radius dp and uses other coordinates. Have I skipped some part which I am not allowed to ? Why is my result wrong?

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Once you have dF at every point, it becomes a classical mechanics problem. Use the integral definition of torque. Since the mass distribution is homogeneous, and the magnetic field is constant, it follows that dF=constant*dm. The force will be proportional to mass, hence the resultant force can be applied at the center of mass, and taking the torque wherever from there will give you a result.

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Because I think it should be (F)(R), instead of multiplying F by 2R. You are taking the torque with respect to the center of mass after all.

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